How to solve this simple equation $frac{46}{y} + y = 25$?












1












$begingroup$


How do I solve this simple equation? $frac{46}{y} + y = 25$



I know that the answer is $2$, but how do I arrive at that?






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
    $endgroup$
    – Wuestenfux
    Jan 5 at 13:19










  • $begingroup$
    Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:20










  • $begingroup$
    $2$ is not the only solution - the quadratic formula will give you another.
    $endgroup$
    – Mark Bennet
    Jan 5 at 13:21






  • 1




    $begingroup$
    What aobut $y=23$?
    $endgroup$
    – Yanko
    Jan 5 at 13:22
















1












$begingroup$


How do I solve this simple equation? $frac{46}{y} + y = 25$



I know that the answer is $2$, but how do I arrive at that?






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
    $endgroup$
    – Wuestenfux
    Jan 5 at 13:19










  • $begingroup$
    Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:20










  • $begingroup$
    $2$ is not the only solution - the quadratic formula will give you another.
    $endgroup$
    – Mark Bennet
    Jan 5 at 13:21






  • 1




    $begingroup$
    What aobut $y=23$?
    $endgroup$
    – Yanko
    Jan 5 at 13:22














1












1








1





$begingroup$


How do I solve this simple equation? $frac{46}{y} + y = 25$



I know that the answer is $2$, but how do I arrive at that?






algebra-precalculus







share|cite|improve this question











$endgroup$




How do I solve this simple equation? $frac{46}{y} + y = 25$



I know that the answer is $2$, but how do I arrive at that?






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 13:28









6005

36.3k751125




36.3k751125










asked Jan 5 at 13:18









brilliantbrilliant

30729




30729








  • 1




    $begingroup$
    Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
    $endgroup$
    – Wuestenfux
    Jan 5 at 13:19










  • $begingroup$
    Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:20










  • $begingroup$
    $2$ is not the only solution - the quadratic formula will give you another.
    $endgroup$
    – Mark Bennet
    Jan 5 at 13:21






  • 1




    $begingroup$
    What aobut $y=23$?
    $endgroup$
    – Yanko
    Jan 5 at 13:22














  • 1




    $begingroup$
    Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
    $endgroup$
    – Wuestenfux
    Jan 5 at 13:19










  • $begingroup$
    Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:20










  • $begingroup$
    $2$ is not the only solution - the quadratic formula will give you another.
    $endgroup$
    – Mark Bennet
    Jan 5 at 13:21






  • 1




    $begingroup$
    What aobut $y=23$?
    $endgroup$
    – Yanko
    Jan 5 at 13:22








1




1




$begingroup$
Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
$endgroup$
– Wuestenfux
Jan 5 at 13:19




$begingroup$
Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
$endgroup$
– Wuestenfux
Jan 5 at 13:19












$begingroup$
Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
$endgroup$
– Thomas Shelby
Jan 5 at 13:20




$begingroup$
Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
$endgroup$
– Thomas Shelby
Jan 5 at 13:20












$begingroup$
$2$ is not the only solution - the quadratic formula will give you another.
$endgroup$
– Mark Bennet
Jan 5 at 13:21




$begingroup$
$2$ is not the only solution - the quadratic formula will give you another.
$endgroup$
– Mark Bennet
Jan 5 at 13:21




1




1




$begingroup$
What aobut $y=23$?
$endgroup$
– Yanko
Jan 5 at 13:22




$begingroup$
What aobut $y=23$?
$endgroup$
– Yanko
Jan 5 at 13:22










4 Answers
4






active

oldest

votes


















3












$begingroup$

$$frac{46}{y}+y = 25$$



Here, assuming $y neq 0$, you can multiply both sides of the equation by $y$, yielding



$$46+y^2 = 25y$$



$$y^2-25y+46 = 0$$



Here, you can solve the quadratic equation by factoring, completing the square, or the Quadratic Formula, but the first way is the simplest, considering $a = 1$:



$$(y+y_1)(y+y_2) = 0$$



The key here is to ask yourself: which two numbers multiply to give $+46$ and add to give $-25$? Clearly the two numbers must be negative, and you can figure out they’re $-23$ and $-2$. So, the factored equation becomes



$$(y-23)(y-2) = 0$$



Setting either factor equal to $0$ yields




  • $$y-23 = 0 iff y = 23$$

  • $$y-2 = 0 iff y = 2$$


You could also use the Quadratic Formula if desired:




$$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$



$$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$



$$y = 23; quad y = 2$$







share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    $$y^2-25y+46=0implies y=2,23$$
    you can use middle term method or the quadratic formula.
    the formula is...
    if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      Since you see $frac{46}{y}$ in there, it's a good idea to multiply both sides by $y$:



      begin{align*}
      left(frac{46}{y} + yright) cdot y &= 25 cdot y \
      46 + y^2 &= 25y
      end{align*}



      Now we bring all the terms over to one side:
      $$
      y^2 - 25y + 46 = 0
      $$
      so you have a quadratic equation, and you can solve it. (Quadratic formula or factoring!)



      If you happen to get $y = 0$ as a solution to the equation, you have to throw it out, since at the beginning there was $frac{46}{y}$, which means that $y$ was not $0$. But here it turns out that $y = 0$ is not a solution.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Multiply both sides by $y$ and you get
        $$46 + y^2 = 25y$$
        $$Leftrightarrow 0 = y^2 -25y + 46$$
        Now calculating the discriminant $D = (-25)^2 -4cdot 46 = 625-184 = 441 = 21^2$ gives
        $$y = frac{25 pm sqrt{21^2}}{2}$$
        So $y = frac{25+21}{2} = 23$ or $y = frac{25-21}{2} = 2$.






        share|cite|improve this answer









        $endgroup$













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          4 Answers
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          active

          oldest

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          4 Answers
          4






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          active

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          active

          oldest

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          3












          $begingroup$

          $$frac{46}{y}+y = 25$$



          Here, assuming $y neq 0$, you can multiply both sides of the equation by $y$, yielding



          $$46+y^2 = 25y$$



          $$y^2-25y+46 = 0$$



          Here, you can solve the quadratic equation by factoring, completing the square, or the Quadratic Formula, but the first way is the simplest, considering $a = 1$:



          $$(y+y_1)(y+y_2) = 0$$



          The key here is to ask yourself: which two numbers multiply to give $+46$ and add to give $-25$? Clearly the two numbers must be negative, and you can figure out they’re $-23$ and $-2$. So, the factored equation becomes



          $$(y-23)(y-2) = 0$$



          Setting either factor equal to $0$ yields




          • $$y-23 = 0 iff y = 23$$

          • $$y-2 = 0 iff y = 2$$


          You could also use the Quadratic Formula if desired:




          $$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$



          $$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$



          $$y = 23; quad y = 2$$







          share|cite|improve this answer











          $endgroup$


















            3












            $begingroup$

            $$frac{46}{y}+y = 25$$



            Here, assuming $y neq 0$, you can multiply both sides of the equation by $y$, yielding



            $$46+y^2 = 25y$$



            $$y^2-25y+46 = 0$$



            Here, you can solve the quadratic equation by factoring, completing the square, or the Quadratic Formula, but the first way is the simplest, considering $a = 1$:



            $$(y+y_1)(y+y_2) = 0$$



            The key here is to ask yourself: which two numbers multiply to give $+46$ and add to give $-25$? Clearly the two numbers must be negative, and you can figure out they’re $-23$ and $-2$. So, the factored equation becomes



            $$(y-23)(y-2) = 0$$



            Setting either factor equal to $0$ yields




            • $$y-23 = 0 iff y = 23$$

            • $$y-2 = 0 iff y = 2$$


            You could also use the Quadratic Formula if desired:




            $$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$



            $$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$



            $$y = 23; quad y = 2$$







            share|cite|improve this answer











            $endgroup$
















              3












              3








              3





              $begingroup$

              $$frac{46}{y}+y = 25$$



              Here, assuming $y neq 0$, you can multiply both sides of the equation by $y$, yielding



              $$46+y^2 = 25y$$



              $$y^2-25y+46 = 0$$



              Here, you can solve the quadratic equation by factoring, completing the square, or the Quadratic Formula, but the first way is the simplest, considering $a = 1$:



              $$(y+y_1)(y+y_2) = 0$$



              The key here is to ask yourself: which two numbers multiply to give $+46$ and add to give $-25$? Clearly the two numbers must be negative, and you can figure out they’re $-23$ and $-2$. So, the factored equation becomes



              $$(y-23)(y-2) = 0$$



              Setting either factor equal to $0$ yields




              • $$y-23 = 0 iff y = 23$$

              • $$y-2 = 0 iff y = 2$$


              You could also use the Quadratic Formula if desired:




              $$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$



              $$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$



              $$y = 23; quad y = 2$$







              share|cite|improve this answer











              $endgroup$



              $$frac{46}{y}+y = 25$$



              Here, assuming $y neq 0$, you can multiply both sides of the equation by $y$, yielding



              $$46+y^2 = 25y$$



              $$y^2-25y+46 = 0$$



              Here, you can solve the quadratic equation by factoring, completing the square, or the Quadratic Formula, but the first way is the simplest, considering $a = 1$:



              $$(y+y_1)(y+y_2) = 0$$



              The key here is to ask yourself: which two numbers multiply to give $+46$ and add to give $-25$? Clearly the two numbers must be negative, and you can figure out they’re $-23$ and $-2$. So, the factored equation becomes



              $$(y-23)(y-2) = 0$$



              Setting either factor equal to $0$ yields




              • $$y-23 = 0 iff y = 23$$

              • $$y-2 = 0 iff y = 2$$


              You could also use the Quadratic Formula if desired:




              $$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$



              $$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$



              $$y = 23; quad y = 2$$








              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 5 at 13:34

























              answered Jan 5 at 13:27









              KM101KM101

              6,0701525




              6,0701525























                  2












                  $begingroup$

                  $$y^2-25y+46=0implies y=2,23$$
                  you can use middle term method or the quadratic formula.
                  the formula is...
                  if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$






                  share|cite|improve this answer











                  $endgroup$


















                    2












                    $begingroup$

                    $$y^2-25y+46=0implies y=2,23$$
                    you can use middle term method or the quadratic formula.
                    the formula is...
                    if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$






                    share|cite|improve this answer











                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      $$y^2-25y+46=0implies y=2,23$$
                      you can use middle term method or the quadratic formula.
                      the formula is...
                      if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$






                      share|cite|improve this answer











                      $endgroup$



                      $$y^2-25y+46=0implies y=2,23$$
                      you can use middle term method or the quadratic formula.
                      the formula is...
                      if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 5 at 13:30

























                      answered Jan 5 at 13:22









                      Rakibul Islam PrinceRakibul Islam Prince

                      988211




                      988211























                          0












                          $begingroup$

                          Since you see $frac{46}{y}$ in there, it's a good idea to multiply both sides by $y$:



                          begin{align*}
                          left(frac{46}{y} + yright) cdot y &= 25 cdot y \
                          46 + y^2 &= 25y
                          end{align*}



                          Now we bring all the terms over to one side:
                          $$
                          y^2 - 25y + 46 = 0
                          $$
                          so you have a quadratic equation, and you can solve it. (Quadratic formula or factoring!)



                          If you happen to get $y = 0$ as a solution to the equation, you have to throw it out, since at the beginning there was $frac{46}{y}$, which means that $y$ was not $0$. But here it turns out that $y = 0$ is not a solution.






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            Since you see $frac{46}{y}$ in there, it's a good idea to multiply both sides by $y$:



                            begin{align*}
                            left(frac{46}{y} + yright) cdot y &= 25 cdot y \
                            46 + y^2 &= 25y
                            end{align*}



                            Now we bring all the terms over to one side:
                            $$
                            y^2 - 25y + 46 = 0
                            $$
                            so you have a quadratic equation, and you can solve it. (Quadratic formula or factoring!)



                            If you happen to get $y = 0$ as a solution to the equation, you have to throw it out, since at the beginning there was $frac{46}{y}$, which means that $y$ was not $0$. But here it turns out that $y = 0$ is not a solution.






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              Since you see $frac{46}{y}$ in there, it's a good idea to multiply both sides by $y$:



                              begin{align*}
                              left(frac{46}{y} + yright) cdot y &= 25 cdot y \
                              46 + y^2 &= 25y
                              end{align*}



                              Now we bring all the terms over to one side:
                              $$
                              y^2 - 25y + 46 = 0
                              $$
                              so you have a quadratic equation, and you can solve it. (Quadratic formula or factoring!)



                              If you happen to get $y = 0$ as a solution to the equation, you have to throw it out, since at the beginning there was $frac{46}{y}$, which means that $y$ was not $0$. But here it turns out that $y = 0$ is not a solution.






                              share|cite|improve this answer









                              $endgroup$



                              Since you see $frac{46}{y}$ in there, it's a good idea to multiply both sides by $y$:



                              begin{align*}
                              left(frac{46}{y} + yright) cdot y &= 25 cdot y \
                              46 + y^2 &= 25y
                              end{align*}



                              Now we bring all the terms over to one side:
                              $$
                              y^2 - 25y + 46 = 0
                              $$
                              so you have a quadratic equation, and you can solve it. (Quadratic formula or factoring!)



                              If you happen to get $y = 0$ as a solution to the equation, you have to throw it out, since at the beginning there was $frac{46}{y}$, which means that $y$ was not $0$. But here it turns out that $y = 0$ is not a solution.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 5 at 13:27









                              60056005

                              36.3k751125




                              36.3k751125























                                  0












                                  $begingroup$

                                  Multiply both sides by $y$ and you get
                                  $$46 + y^2 = 25y$$
                                  $$Leftrightarrow 0 = y^2 -25y + 46$$
                                  Now calculating the discriminant $D = (-25)^2 -4cdot 46 = 625-184 = 441 = 21^2$ gives
                                  $$y = frac{25 pm sqrt{21^2}}{2}$$
                                  So $y = frac{25+21}{2} = 23$ or $y = frac{25-21}{2} = 2$.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Multiply both sides by $y$ and you get
                                    $$46 + y^2 = 25y$$
                                    $$Leftrightarrow 0 = y^2 -25y + 46$$
                                    Now calculating the discriminant $D = (-25)^2 -4cdot 46 = 625-184 = 441 = 21^2$ gives
                                    $$y = frac{25 pm sqrt{21^2}}{2}$$
                                    So $y = frac{25+21}{2} = 23$ or $y = frac{25-21}{2} = 2$.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Multiply both sides by $y$ and you get
                                      $$46 + y^2 = 25y$$
                                      $$Leftrightarrow 0 = y^2 -25y + 46$$
                                      Now calculating the discriminant $D = (-25)^2 -4cdot 46 = 625-184 = 441 = 21^2$ gives
                                      $$y = frac{25 pm sqrt{21^2}}{2}$$
                                      So $y = frac{25+21}{2} = 23$ or $y = frac{25-21}{2} = 2$.






                                      share|cite|improve this answer









                                      $endgroup$



                                      Multiply both sides by $y$ and you get
                                      $$46 + y^2 = 25y$$
                                      $$Leftrightarrow 0 = y^2 -25y + 46$$
                                      Now calculating the discriminant $D = (-25)^2 -4cdot 46 = 625-184 = 441 = 21^2$ gives
                                      $$y = frac{25 pm sqrt{21^2}}{2}$$
                                      So $y = frac{25+21}{2} = 23$ or $y = frac{25-21}{2} = 2$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 5 at 13:29









                                      Jonas De SchouwerJonas De Schouwer

                                      3768




                                      3768






























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                                          Mont Emei

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                                          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
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